# Thread: why am i blanking so hardcore? -stats

1. ## why am i blanking so hardcore? -stats

let X(bar) be the mean of a random sample of size n=10 from a distribution with pdf f(x) = 6x(1-x) for 0<x<1. Find the mean and the variance of X(bar).

I feel like there is just a simple formula for this but I cannot for the life of me remember this. It sounds like a foreign language right now! Please help with any input you can, this is due tomorrow.

2. Originally Posted by Dubulus
let X(bar) be the mean of a random sample of size n=10 from a distribution with pdf f(x) = 6x(1-x) for 0<x<1. Find the mean and the variance of X(bar).

I feel like there is just a simple formula for this but I cannot for the life of me remember this. It sounds like a foreign language right now! Please help with any input you can, this is due tomorrow.
$\displaystyle \overline{x} = \frac{X_1 + X_2 + \, .... \, + X_9 + X_{10}}{10}$

$\displaystyle \Rightarrow E(\overline{x}) = E\left( \frac{X_1 + X_2 + \, .... \, + X_9 + X_{10}}{10}\right)$ $\displaystyle = \frac{E(X_1) + E(X_2) + \, .... \, + E(X_9) + E(X_{10})}{10} = E(X)$.

And you calculate $\displaystyle E(X)$ from the given pdf of X.

Use the same approach to find $\displaystyle E\left(\overline{x}^2\right)$. Then get $\displaystyle Var(\overline{x})$ in the usual way.

3. wont we divide by (n-1) which is 9 in this case? (something to do with degrees of freedom cos this is a sample that we need)

4. Originally Posted by champrock
wont we divide by (n-1) which is 9 in this case? (something to do with degrees of freedom cos this is a sample that we need)
When you find a mean of a sample you divide by the sample size. Unless you know something about this question that you're not telling anyone ....